Step by step
The question
Question: for which k do 2x + 3y = 0 and 6x + ky = 0 have infinitely many solutions?
We call them ① and ②.
Infinitely many solutions means every point on a line works for both.
1 Make the x parts the same
To compare the equations, we make the parts of one letter the same.
k is with y, so we use the other letter, x: it has plain numbers in front.
The x parts: 2x in ① and 6x in ②.
6 divided by 2 is 3.
So 2x times 3 is 6x.
We multiply every part of ① by 3.
We do it on both sides, so they stay equal.
3 times 2x is 6x.
3 times 3y is 9y.
3 times 0 is 0.
We call it ③.
2 Compare ③ and ②
We write ② again, under ③.
If the left sides are the same, both equations talk about the same amount, 6x + 9y.
One amount cannot be two different numbers at once.
So no point works: no solution.
If the numbers are the same too, it is one equation written twice.
Every point on that line works: infinitely many solutions.
If the left sides are different, we can take one equation away from the other.
Some y is left, so we can find y.
That gives one solution.
Careful!
We compare with ③, not ①.
① was times 3.
3y times 3 is 9y.
3 Find k
We want infinitely many solutions.
So the left sides must be the same: ky = 9y.
k times y is 9 times y when k is 9.
So k = 9.
And the numbers, 0 and 0, are the same too.
Good.
4 Check
Let's check.
We put k = 9 into ②.
Take ② away from ③.
The left sides are the same, so together they make 0.
The numbers: 0 take away 0.
0 = 0 is always true.
They are the same equation.
It works!
On a grid, both equations give the same line.
② lies right on top of ①.
Every point on the line works for both: infinitely many solutions.
Let's look back.
First: make the x parts the same.
Then: compare ③ and ②.
Then: find k.
Last: check.